A Jar contains 18 balls. 3 blue balls are removed from the jar and not replaced.Now the probability of getting a blue ball is 1/5 then how many blue balls jar contains initially?

• x/15 = 1/5
  x =3
  3+3(removed 3 blue balls) = 6
• 10 years agoHelpfull: Yes(67) No(0)
• Let initially no. of blue balls are x.
  
  so, initially probability of getting 1 blue ball = (x) combination 1 = x
  after remove 3 blue balls no. of blue balls in the jar = x-3
  now the probability of getting 1 blue ball = (x-3) combination 1 = x-3
  as, x-3 = 1/5 , x = 3.
  so, initially no. of blue balls in the jar = 3 + 3 = 6. (because after remove 3 blue balls we get 3 blue balls in the jar.)
• 10 years agoHelpfull: Yes(10) No(7)
• Let left blue ball is x,then a/c
  xc1/15c1=1/5
  thus, x=3
  so total no. of blue ball is 3+3=6.
  
• 10 years agoHelpfull: Yes(10) No(1)
• 6 blue balls.
  
• 10 years agoHelpfull: Yes(3) No(0)
• more than 3 and less than 6
• 10 years agoHelpfull: Yes(2) No(0)
• initially blue balls=x
  3 blue balls are removed
  so, x-3/15=1/5
  x-3=3
  x=6(ans)
  
• 10 years agoHelpfull: Yes(2) No(0)
• after removing 3 ball remaining ball=15
  p(a)=n(a)/n(s)=1/5; ow n(s)=15
  n(a)=3:
  so the total blue ball=3+3=6
• 10 years agoHelpfull: Yes(1) No(0)
• let the total number of blue balls = x.
  now, out of 18 balls, three blue balls are removed, so the number of balls will be 18-3 = 15
  
  we have (x-3)/15 = 1/5 [the probability of blue balls that are presented in the jar]
  ==> x-3 = 3
  x=6
  the number of blue balls the jar contained initially = 6
• 6 years agoHelpfull: Yes(1) No(0)
• 8 blue balls
• 10 years agoHelpfull: Yes(0) No(8)
• 6
  at first 3blue balls are removed then remaining balls=18-3=15
  p=xc1/15c1=1/5;
  x=3
  
• 10 years agoHelpfull: Yes(0) No(0)
• x/15=1/5
  x=3
• 10 years agoHelpfull: Yes(0) No(1)
• 6 blue balls
  
• 10 years agoHelpfull: Yes(0) No(1)
• x/15 = 1/5
  x =3
  3+3 = 6
  
• 10 years agoHelpfull: Yes(0) No(0)
• Here are not mentioned fully blue balls
• 8 years agoHelpfull: Yes(0) No(1)